HDU 1710 Binary Tree Traversals

Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6914    Accepted Submission(s): 3194

Problem Description

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.

In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.

In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.

In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.

Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.

 

Input

The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.

 

Output

For each test case print a single line specifying the corresponding postorder sequence.

 

Sample Input

91 2 4 7 3 5 8 9 64 7 2 1 8 5 9 3 6

 

Sample Output

7 4 2 8 9 5 6 3 1

解题思路：题目比较简单，题目给出树的前序遍历和中序遍历，然后要求你输出后序遍历。根据前序和中序利用递归的方法把数建好，然后利用队列和递归输出后序遍历。。。

代码如下：

#include<cstdio>#include<math.h>#include<cstring>#include<queue>#include<algorithm>#define maxn 1001using namespace std;int in[maxn],pre[maxn],l[maxn],r[maxn];//pre表示前序遍历序列，in表示中序遍历序列int f(int inl,int inr,int prel,int prer){    if(inl>inr) return 0;    int s=pre[prel];//结点    int t=inl;    while(s!=in[t])        t++;    int p=t-inl;    l[s]=f(inl,t-1,prel+1,prel+p);    r[s]=f(t+1,inr,prel+1+p,prer);    return s;}void print(int s){    queue<int>Q;    Q.push(s);    if(!Q.empty())    {        int num=Q.front();        if(l[num]) print(l[num]);        if(r[num]) print(r[num]);        if(num==pre[0]) printf("%d\n",num);        else printf("%d ",num);        Q.pop();    }}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        memset(l,0,sizeof(l));        memset(r,0,sizeof(r));        for(int i=0; i<n; i++)            scanf("%d",&pre[i]);        for(int i=0; i<n; i++)            scanf("%d",&in[i]);        f(0,n-1,0,n-1);        print(pre[0]);    }    return 0;}